(2a+3)x^2+(a+2)x-(3a+5)=0
用十字相乘法对左边分解因式：
{ (2a+3)x+(3a+5) } { x-1}=0
x1=-(3a+5)/(2a+3)
x2=1
∵ x1=-(3a+5)/(2a+3)
=-3/2 (a+5/3)/(a+3/2)
=-3/2(a+3/2+1/6)/(a+3/2)
=-3/2{1+1/[6(a+3/2)}
∵a＞0
∴{1+1/[6(a+3/2)}＞0
∴x1=-3/2{1+1/[6(a+3/2)}＜0
∴较大的一个实数根等于1